Automorphism group of elliptic curves. I am reading Vakil's Algebraic geometry. There is exercise:

Suppose $(\mathrm{E},\mathrm{p})$ is an elliptic curve over an algebraically closed
  field $k$ of characteristic not $2$ or $3$. Show that the automorphism group of $(\mathrm{E},\mathrm{p})$ 
  is isomorphic to $\mathbb{Z}/2$, $\mathbb{Z}/4$, or $\mathbb{Z}/6$.

However, I think it is not correct. I know that the order automprphism group order could be $2,4$ or $6$. But I think it should be isomorphic with $\mathbb{Z}/2\times \mathbb{Z}/2$, and $\mathbb{Z}/2\times \mathbb{Z}/3$. Am I right? Also I think when char  $k=3$, it could be isomorphic with $\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/3$, is it right? Last question, if char $k=2$, how could we get the automprphism group? Since only when char $k\not=2$, we could get the group in term of one fix point $\infty$, and three other point on $\mathbb{P}^{1}$. I have no idea how to compute the group when char $k=2$. More generally, in the theory of elliptic curves, we always need to assume char $k\not =2$, but what should we do if char $k=2$?
 A: In the case that the base field is not of characteristic $2$ or $3$, the curve can be put in the form $$ y^2 = x^3+Ax+B$$ and the automorphisms are of the form $x=u^2x'$ and $y=u^3y'$ where $u^{4}A=A$ and $u^{6}B=B$. If $AB\neq 0$, then $u^{2}=1$ and the group is cyclic of degree 2. If $AB=0$, then either $B=0$ and $u^4=1$ or $A=0$ and $u^6=1$, and in this case there's a natural cyclic group structure on the solutions. (Further, $\Bbb Z/2\times \Bbb Z/3\cong \Bbb Z/6$, so there's no disagreement with that specific part of your idea.)
For characteristics $2,3$, the general idea is that you have to be a little more careful about things. Here, that means how you put the curve in to a canonical form like in the first paragraph, but it can still be done. See for instance appendix A in Silverman's Arithmetic of Elliptic Curves, where automorphism groups are explained for curves in characteristic $2$ and $3$. In short, the group in characteristic 3 can be of order $12$ and the group in characteristic $2$ can be of order $24$. If you go look at the calculations, you should be able to tell whether they're cyclic or not.
